Termination w.r.t. Q proof of /tmp/tmpRL4EnH/z120.srs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q restricted rewrite system:
The TRS R consists of the following rules:

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

C(c(c(a(x1)))) → D(x1)
C(c(c(a(x1)))) → D(d(x1))
B(c(x1)) → B(a(c(x1)))
D(x1) → C(x1)
D(x1) → B(c(x1))
D(b(x1)) → C(x1)
D(b(x1)) → C(c(x1))

The TRS R consists of the following rules:

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

C(c(c(a(x1)))) → D(x1)
C(c(c(a(x1)))) → D(d(x1))
B(c(x1)) → B(a(c(x1)))
D(x1) → C(x1)
D(x1) → B(c(x1))
D(b(x1)) → C(x1)
D(b(x1)) → C(c(x1))

The TRS R consists of the following rules:

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

C(c(c(a(x1)))) → D(x1)
C(c(c(a(x1)))) → D(d(x1))
D(x1) → C(x1)
D(b(x1)) → C(x1)
D(b(x1)) → C(c(x1))

The TRS R consists of the following rules:

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

C(c(c(a(x1)))) → D(x1)
D(x1) → C(x1)
D(b(x1)) → C(x1)

The remaining pairs can at least be oriented weakly.

C(c(c(a(x1)))) → D(d(x1))
D(b(x1)) → C(c(x1))

Used ordering: Polynomial interpretation [25]:

POL(C(x₁)) = 4·x₁
POL(D(x₁)) = 12 + 4·x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = 3 + x₁
POL(c(x₁)) = 6 + x₁
POL(d(x₁)) = 9 + x₁

The following usable rules [17] were oriented:

c(x1) → a(a(x1))
b(c(x1)) → b(a(c(x1)))
c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
d(x1) → b(c(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

C(c(c(a(x1)))) → D(d(x1))
D(b(x1)) → C(c(x1))

The TRS R consists of the following rules:

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(c(c(a(x1)))) → D(d(x1)) at position [0] we obtained the following new rules:

C(c(c(a(b(x0))))) → D(c(c(x0)))
C(c(c(a(x0)))) → D(b(c(x0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

C(c(c(a(b(x0))))) → D(c(c(x0)))
C(c(c(a(x0)))) → D(b(c(x0)))
D(b(x1)) → C(c(x1))

The TRS R consists of the following rules:

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q restricted rewrite system:
The TRS R consists of the following rules:

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))
C(c(c(a(b(x0))))) → D(c(c(x0)))
C(c(c(a(x0)))) → D(b(c(x0)))
D(b(x1)) → C(c(x1))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))
C(c(c(a(b(x0))))) → D(c(c(x0)))
C(c(c(a(x0)))) → D(b(c(x0)))
D(b(x1)) → C(c(x1))

The set Q is empty.
We have obtained the following QTRS:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q restricted rewrite system:
The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

The set Q is empty.
We have obtained the following QTRS:

c(c(c(a(x)))) → d(d(x))
d(b(x)) → c(c(x))
b(c(x)) → b(a(c(x)))
c(x) → a(a(x))
d(x) → b(c(x))
C(c(c(a(b(x))))) → D(c(c(x)))
C(c(c(a(x)))) → D(b(c(x)))
D(b(x)) → C(c(x))

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q restricted rewrite system:
The TRS R consists of the following rules:

c(c(c(a(x)))) → d(d(x))
d(b(x)) → c(c(x))
b(c(x)) → b(a(c(x)))
c(x) → a(a(x))
d(x) → b(c(x))
C(c(c(a(b(x))))) → D(c(c(x)))
C(c(c(a(x)))) → D(b(c(x)))
D(b(x)) → C(c(x))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

The set Q is empty.
We have obtained the following QTRS:

c(c(c(a(x)))) → d(d(x))
d(b(x)) → c(c(x))
b(c(x)) → b(a(c(x)))
c(x) → a(a(x))
d(x) → b(c(x))
C(c(c(a(b(x))))) → D(c(c(x)))
C(c(c(a(x)))) → D(b(c(x)))
D(b(x)) → C(c(x))

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                            ↳ QTRS

                          ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q restricted rewrite system:
The TRS R consists of the following rules:

c(c(c(a(x)))) → d(d(x))
d(b(x)) → c(c(x))
b(c(x)) → b(a(c(x)))
c(x) → a(a(x))
d(x) → b(c(x))
C(c(c(a(b(x))))) → D(c(c(x)))
C(c(c(a(x)))) → D(b(c(x)))
D(b(x)) → C(c(x))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

A(c(c(c(x)))) → D¹(x)
C¹(b(x)) → A(b(x))
B(d(x)) → C¹(x)
D¹(x) → C¹(b(x))
A(c(c(C(x)))) → B(D(x))
A(c(c(C(x)))) → C¹(b(D(x)))
C¹(b(x)) → C¹(a(b(x)))
D¹(x) → B(x)
B(d(x)) → C¹(c(x))
A(c(c(c(x)))) → D¹(d(x))
B(a(c(c(C(x))))) → C¹(c(D(x)))
C¹(x) → A(x)
B(D(x)) → C¹(C(x))
C¹(x) → A(a(x))
B(a(c(c(C(x))))) → C¹(D(x))

The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

A(c(c(c(x)))) → D¹(x)
C¹(b(x)) → A(b(x))
B(d(x)) → C¹(x)
D¹(x) → C¹(b(x))
A(c(c(C(x)))) → B(D(x))
A(c(c(C(x)))) → C¹(b(D(x)))
C¹(b(x)) → C¹(a(b(x)))
D¹(x) → B(x)
B(d(x)) → C¹(c(x))
A(c(c(c(x)))) → D¹(d(x))
B(a(c(c(C(x))))) → C¹(c(D(x)))
C¹(x) → A(x)
B(D(x)) → C¹(C(x))
C¹(x) → A(a(x))
B(a(c(c(C(x))))) → C¹(D(x))

The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

B(a(c(c(C(x))))) → C¹(c(D(x)))
B(D(x)) → C¹(C(x))
B(a(c(c(C(x))))) → C¹(D(x))

The remaining pairs can at least be oriented weakly.

A(c(c(c(x)))) → D¹(x)
C¹(b(x)) → A(b(x))
B(d(x)) → C¹(x)
D¹(x) → C¹(b(x))
A(c(c(C(x)))) → B(D(x))
A(c(c(C(x)))) → C¹(b(D(x)))
C¹(b(x)) → C¹(a(b(x)))
D¹(x) → B(x)
B(d(x)) → C¹(c(x))
A(c(c(c(x)))) → D¹(d(x))
C¹(x) → A(x)
C¹(x) → A(a(x))

Used ordering: Polynomial interpretation [25]:

POL(A(x₁)) = x₁
POL(B(x₁)) = 8·x₁
POL(C(x₁)) = 1
POL(C¹(x₁)) = 4·x₁
POL(D(x₁)) = 2
POL(D¹(x₁)) = 8·x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = 2·x₁
POL(c(x₁)) = 4·x₁
POL(d(x₁)) = 8·x₁

The following usable rules [17] were oriented:

b(D(x)) → c(C(x))
b(a(c(c(C(x))))) → c(c(D(x)))
c(b(x)) → c(a(b(x)))
a(c(c(C(x)))) → c(b(D(x)))
c(x) → a(a(x))
a(c(c(c(x)))) → d(d(x))
d(x) → c(b(x))
b(d(x)) → c(c(x))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ QDPOrderProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

B(d(x)) → C¹(c(x))
A(c(c(c(x)))) → D¹(x)
A(c(c(c(x)))) → D¹(d(x))
C¹(b(x)) → A(b(x))
B(d(x)) → C¹(x)
C¹(x) → A(x)
C¹(x) → A(a(x))
D¹(x) → C¹(b(x))
A(c(c(C(x)))) → C¹(b(D(x)))
A(c(c(C(x)))) → B(D(x))
D¹(x) → B(x)
C¹(b(x)) → C¹(a(b(x)))

The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ QDPOrderProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

A(c(c(c(x)))) → D¹(x)
B(d(x)) → C¹(c(x))
C¹(b(x)) → A(b(x))
A(c(c(c(x)))) → D¹(d(x))
B(d(x)) → C¹(x)
C¹(x) → A(x)
C¹(x) → A(a(x))
D¹(x) → C¹(b(x))
A(c(c(C(x)))) → C¹(b(D(x)))
D¹(x) → B(x)
C¹(b(x)) → C¹(a(b(x)))

The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

A(c(c(C(x)))) → C¹(b(D(x)))

The remaining pairs can at least be oriented weakly.

A(c(c(c(x)))) → D¹(x)
B(d(x)) → C¹(c(x))
C¹(b(x)) → A(b(x))
A(c(c(c(x)))) → D¹(d(x))
B(d(x)) → C¹(x)
C¹(x) → A(x)
C¹(x) → A(a(x))
D¹(x) → C¹(b(x))
D¹(x) → B(x)
C¹(b(x)) → C¹(a(b(x)))

Used ordering: Polynomial interpretation [25]:

POL(A(x₁)) = x₁
POL(B(x₁)) = x₁
POL(C(x₁)) = 2
POL(C¹(x₁)) = x₁
POL(D(x₁)) = 4
POL(D¹(x₁)) = 2·x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = 2·x₁
POL(c(x₁)) = 4·x₁
POL(d(x₁)) = 8·x₁

The following usable rules [17] were oriented:

b(D(x)) → c(C(x))
b(a(c(c(C(x))))) → c(c(D(x)))
c(b(x)) → c(a(b(x)))
a(c(c(C(x)))) → c(b(D(x)))
c(x) → a(a(x))
a(c(c(c(x)))) → d(d(x))
d(x) → c(b(x))
b(d(x)) → c(c(x))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ QDPOrderProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

B(d(x)) → C¹(c(x))
A(c(c(c(x)))) → D¹(x)
A(c(c(c(x)))) → D¹(d(x))
C¹(b(x)) → A(b(x))
B(d(x)) → C¹(x)
C¹(x) → A(x)
D¹(x) → C¹(b(x))
C¹(x) → A(a(x))
C¹(b(x)) → C¹(a(b(x)))
D¹(x) → B(x)

The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

B(d(x)) → C¹(c(x))
A(c(c(c(x)))) → D¹(x)
A(c(c(c(x)))) → D¹(d(x))
B(d(x)) → C¹(x)
D¹(x) → C¹(b(x))
D¹(x) → B(x)

The remaining pairs can at least be oriented weakly.

C¹(b(x)) → A(b(x))
C¹(x) → A(x)
C¹(x) → A(a(x))
C¹(b(x)) → C¹(a(b(x)))

Used ordering: Polynomial interpretation [25]:

POL(A(x₁)) = x₁
POL(B(x₁)) = x₁
POL(C(x₁)) = 2·x₁
POL(C¹(x₁)) = x₁
POL(D(x₁)) = 1 + 2·x₁
POL(D¹(x₁)) = 2 + x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = 1 + x₁
POL(c(x₁)) = 2 + x₁
POL(d(x₁)) = 3 + x₁

The following usable rules [17] were oriented:

b(D(x)) → c(C(x))
b(a(c(c(C(x))))) → c(c(D(x)))
c(b(x)) → c(a(b(x)))
a(c(c(C(x)))) → c(b(D(x)))
c(x) → a(a(x))
a(c(c(c(x)))) → d(d(x))
d(x) → c(b(x))
b(d(x)) → c(c(x))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ QDPOrderProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

                                            ↳ QDP

                                              ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

C¹(b(x)) → A(b(x))
C¹(x) → A(x)
C¹(x) → A(a(x))
C¹(b(x)) → C¹(a(b(x)))

The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 4 less nodes.
We have reversed the following QTRS:
The set of rules R is

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

The set Q is empty.
We have obtained the following QTRS:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

Q restricted rewrite system:
The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

The set Q is empty.
We have obtained the following QTRS:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

Q restricted rewrite system:
The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))

Q is empty.